An $A C$ source is connected with a resistance $(R)$ and an uncharged capacitance $(C)$ , in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at $t = 0$ , emf is zero)

\frac{π}{4 ω}

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\frac{2 π}{ω}

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\frac{π}{2 ω}

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\frac{3 π}{2 ω}

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Solution

The correct option is A $\frac{π}{4 ω}$
The positions of the phasors at t=0 is shown in the diagram on the left. $V_{R}$ is in phase with the current while $V_{C}$ lags the current by $\frac{π}{2}$ . For $V_{R}$ to be in phase with the initial phasor of $V_{C}$ , the $V_{R}$ phasor has to be rotated counterclockwise by an angle $\frac{3 π}{2}$ . With angular speed ω, if this angle is traced in time t then
$\begin{matrix} ω t = \frac{3 π}{2} o r, t = \frac{3 π}{2 ω} \end{matrix}$

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